# Why is $A \leq_T \bar{A}$?

I know there's some easy proof for this, but why is $A \leq_T \bar{A}$ ? ... Suppose that $\bar{A}$ is c.e. and could loop, does this mean that the oracle $TM$ $M^{\bar{A}}$ could also loop?

But shouldn't $\leq_T$ mean that $M^{\bar{A}}$ be decidable?

• Oracles don't loop, that's the point of oracles: they always give you the correct answer. – Vladimir Lysikov Sep 20 '17 at 11:20
• oh right, would be handy in my exam – Link L Sep 20 '17 at 12:23

A language $A$ Turing-reduces to $\overline{A}$ since a Turing machine with an $\overline{A}$-oracle can compute $A$. On input $x$, it determines whether $x \in \overline{A}$ using the oracle, and then outputs the reverse answer.
Your misunderstanding is in the definition of an oracle. An oracle is a hypothetical device that answers the question "Is string $x$ in language $L$?" in a single computational step. It doesn't matter whether or not an actual Turing machine could answer that question; the oracle is defined to answer it, and answer it immediately.